An alkene on reductive as well as oxidative ozonolysis gave only one product. The alkene is (a) propene (b) 2,3-dimethyl-2-butene (c) pent-2-ene (d) 2 -methyl propene

Short Answer

Expert verified
Possible revised answer choices: (a) Propene (b) 2,3-dimethyl-2-butene (c) 2-methyl propene Correct Answer: All of the above (a, b, and c)

Step by step solution

01

Understand ozonolysis reaction

In ozonolysis, ozone (O3) breaks the double bond in the alkene, cleaving the molecule and forming two smaller molecules. In the two types of ozonolysis, reductive ozonolysis (using reducing agents such as DMS or Zn) and oxidative ozonolysis (using oxidizing agents such as H2O2), products will be different if there are allylic positions (carbon-carbon single bonds adjacent to the carbon-carbon double bond) present in the molecule. Only one product will be obtained if the molecule does not have any allylic positions.
02

Analyze each option

We'll look at the possible ozonolysis products for each alkene option. (a) Propene: Upon ozonolysis, propene will form an aldehyde (acetaldehyde) and a carbonyl compound (formaldehyde). Since both the reductive and oxidative ozonolysis form the same product, propene is a possible answer. (b) 2,3-dimethyl-2-butene: Ozonolysis of this alkene will form an aldehyde (isobutyraldehyde) and a ketone (acetone). Since there are no allylic positions, both the reductive and oxidative ozonolysis form the same product, making 2,3-dimethyl-2-butene a possible answer as well. (c) Pent-2-ene: Ozonolysis of this alkene will form an aldehyde (propanal) and a ketone (acetone). However, the allylic positions in the molecule allow for different products in reductive and oxidative ozonolysis, so it does not yield only one product. (d) 2-methyl propene: Ozonolysis of this alkene will form a ketone (acetone) and a carbonyl compound (formaldehyde). Since there are no allylic positions, both the reductive and oxidative ozonolysis form the same product, making 2-methyl propene a possible option as well. Although there are three viable options (propene, 2,3-dimethyl-2-butene, and 2-methyl propene), the question seems to have an error, possibly providing a list of alkenes that may yield only one identical product in both reductive and oxidative ozonolysis. In this case, the answer choices should be revised accordingly.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

\(\mathrm{H}_{3} \mathrm{C}-\mathrm{C} \equiv \mathrm{CH}+\mathrm{CH}_{3} \mathrm{MgI} \longrightarrow(\mathrm{A})+\mathrm{CH}_{4}\) \((\mathrm{A})+\mathrm{CH}_{3} \mathrm{I} \longrightarrow(\mathrm{B})+\mathrm{MgI}_{2}\) (A) and (B) are (a) \(\mathrm{H}-\mathrm{C} \equiv \mathrm{C}-\mathrm{MgI}\) and \(\mathrm{CH}_{4}\) (b) \(\mathrm{H}_{3} \mathrm{C}-\mathrm{C} \equiv \mathrm{C}-\mathrm{CH}_{3}\) and \(\mathrm{CH}_{4}\) (c) \(\mathrm{CH}_{3}-\mathrm{C} \equiv \mathrm{C}-\mathrm{MgI}\) and \(\mathrm{CH}_{3}-\mathrm{C} \equiv \mathrm{C}-\mathrm{CH}_{3}\) (d) \(\mathrm{H}_{3} \mathrm{C}-\mathrm{C} \equiv \mathrm{C}-\mathrm{I}\) and

A hydrocarbon \(\mathrm{C}_{5} \mathrm{H}_{12}\) (A) on treatment with chlorine undergoes monochlorination by the replacement of the only \(1^{\circ}\) hydrogen to give(B) \(\mathrm{C}_{5} \mathrm{H}_{11} \mathrm{Cl}\). When (B) is treated with Na metal in ether, the product obtained could be (a) \(\quad\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CH}-\mathrm{CH}\left(\mathrm{CH}_{3}\right)_{2}\) (b) \(\mathrm{CH}_{3}-\mathrm{C}\left(\mathrm{CH}_{3}\right)_{2}-\mathrm{C}\left(\mathrm{CH}_{3}\right)_{2}-\mathrm{CH}_{3}\) (c) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{C}\left(\mathrm{CH}_{3}\right)_{3}\) (d) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{CH}_{2} \mathrm{Cl}\)

\(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CuLi}\) is known as (a) Corey-House reagent (b) Gilman reagent (c) Grignard reagent (d) Ullmann reagent

The conversion can be successfully carried out by (a) \(\mathrm{H}^{+} / \mathrm{H}_{2} \mathrm{O}\) (b) Hydroboration-oxidation (c) HBr addition in the presence of peroxides followed by hydrolysis with aqueous KOH (d) Oxymercuration-reduction

Directions: This section contains 1 paragraph. Based upon the paragraph, 3 multiple choice questions have to be answered. Each question has 4 choices (a), (b), (c) and (d), out of which ONLY ONE is correct. Passage I Alkenes undergo a variety of reactions like hydrogenation, halogenation, oxidation, epoxidation and so on. The double bond of the alkenes acts as electron donor. An unsaturated hydrocarbon gave, on treatment with alkaline \(\mathrm{KMnO}_{4}\) under hot conditions, one mole of adipic acid. The unsaturated hydrocarbon could be (a) \(\mathrm{H}_{3} \mathrm{C}-\mathrm{C} \equiv \mathrm{C}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{CH}_{3}\) (b) \(\mathrm{H}_{3} \mathrm{C}-\mathrm{CH}_{2}-\mathrm{C} \equiv \mathrm{C}-\mathrm{CH}_{2}-\mathrm{CH}_{3}\)

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free